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Is the profinite completion of $SL_2(\mathbb{Z})$ equal to $SL_2(\hat{\mathbb{Z}})$?

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up vote 4 down vote accepted

No. The congruence kernel of the natural epimorphism $\widehat{\mathrm{SL}_2(\mathbb{Z})} \to \mathrm{SL}_2(\widehat{\mathbb{Z}})$ is a free profinite group of countably infinite rank. This is Theorem 8.8.1. of L. Ribes, P. Zalesskii, Profinite groups, Springer-Verlag, 2010.

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